Polar Partial Derivatives - Boas Ch 4 Sect 1 Prob 13

[rhdinah]

Homework Statement

If $z=x^2+2y^2$, find the following partial derivative:
\Big(\frac{∂z}{∂\theta}\Big)_x

Homework Equations

$x=r cos(\theta), ~y=r sin(\theta),~r^2=x^2+y^2,~\theta=tan^{-1}\frac{y}{x}$

The Attempt at a Solution

I've been using Boas for self-study and been working on this problem for almost a week now and cannot get the expected answer. I'm persistent though and have decided to look outside my study area for some help. So if you can add some direction or pointers I'd appreciate it.

The first thing we need to do is to simplify the original equation since keeping $x$ constant means the $x^2$ term is a constant and will differentiate to 0 ... so $z=2y^2$.
\frac{∂z}{∂\theta}=\frac{∂z}{∂y}\frac{∂y}{∂\theta}=4 r sin(\theta)*r cos(\theta)=4 r^2sin(\theta)cos(\theta)
Now the expected answer is $4 r^2 tan(\theta)$ and the apparent only way to get to the expected answer from here is to divide by $cos^2(\theta)$ ... which is working backwards of course. I've done and redone this work from a number of approaches and have gotten the same answer. I just don't see where this $tan$ function is coming from.

One of my problems is that I don't know how to do the special case of partial derivatives on Wolfram|Alpha ... while writing standard partials is easy, setting the constraint on what stays constant is difficult to conjure. Plus the engine doesn't convert easily from polar to cartesian well. So if there are some hints there ... so far, I've not found them on my googling of the net.

I've had no problems with any other of the partials, plus all the exercises I've found on the Internet have been straight-forward. So I wonder what I'm doing wrong here. Her next problem 14, $\Big(\frac{∂z}{∂\theta}\Big)_y$, goes down a similar path so knowing how to approach this problem will help with the next. Thanks!

 

[fresh_42]

Why don't you write $z=z(\theta)$?

 

[rhdinah]

Thank you! Okay, let's give it a go ... been thinking about this all day ... :-)

$z=x^2+2y^2=(x^2+y^2)+y^2=r^2+r^2sin^2(\theta)=r^2(1+sin^2(\theta))$

$\Big(\frac{∂z}{∂\theta}\Big)=\Big(\frac{∂}{∂\theta}\Big)r^2(1+sin^2(\theta))=2r^2sin(\theta)cos(\theta)=r^2sin(2\theta)$

Now by *not* collecting terms and dropping $x^2$ as before we get the $4r^2sin(\theta)cos(\theta)=2r^2 sin(2\theta)$ as before ... very confusing ... any other pointers please? Thanks!

 

[fresh_42]

Thank you! Okay, let's give it a go ... been think about this all day ... :-)

$z=x^2+2y^2=(x^2+y^2)+y^2=r^2+r^2sin^2(\theta)=r^2(1+sin^2(\theta))$

$\Big(\frac{∂z}{∂\theta}\Big)=\Big(\frac{∂}{∂\theta}\Big)r^2(1+sin^2(\theta))=2r^2sin(\theta)cos(\theta)=r^2sin(2\theta)$

Now by *not* collecting terms and dropping $x^2$ as before we get the $4r^2sin(\theta)cos(\theta)=2r^2 sin(2\theta)$ as before ... very confusing ... any other pointers please? Thanks!

I had also $\frac{\partial}{\partial \theta}z = 2xy$ and not the factor $4$ as claimed. And even less the tangent of $\theta$.

 

[rhdinah]

Thank you Mr. Fresh_42 for your help! I'm not sure where the $x$ in your solution would have come from as $x$ is supposed to be held constant in the differentiation. Perhaps as a side effect of the $y$ partial. I'm going to try to ask for help from Mary Boas's son as he is monitoring the book for errors and such. If I find some more information I'll post it here. I don't want to be held back so early in the chapter from a seemingly easy problem. We know it may be easy, but it is not trivial. Clearly there is something I'm not seeing here. I'm pretty sure it is not an error in her book as it would have been pointed out a long time ago. Thanks again.

 

[rhdinah]

Professor Boas was kind enough to give me some pointers on this:

$$z=x^2+2y^2=x^2+2(r^2-x^2)=x^2+2(x^2sec^2\theta-x^2)=x^2-2x^2+2x^2sec^2\theta=-x^2+2x^2sec^2\theta$$

$$\frac{d}{d\theta}(-x^2+2x^2sec^2\theta)=4x^2tan\theta sec^2\theta=4 \frac{x^2}{cos^2\theta}tan\theta=4r^2tan\theta\hspace{10 mm} \square$$

 

[fresh_42]

Professor Boas was kind enough to give me some pointers on this:

$$z=x^2+2y^2=x^2+2(r^2-x^2)=x^2+2(x^2sec^2\theta-x^2)=x^2-2x^2+2x^2sec^2\theta=-x^2+2x^2sec^2\theta$$

$$\frac{d}{d\theta}(-x^2+2x^2sec^2\theta)=4x^2tan\theta sec^2\theta=4 \frac{x^2}{cos^2\theta}tan\theta=4r^2tan\theta\hspace{10 mm} \square$$

Do you see the difference between the two solutions and why we got a different result?
In other words: What can be learned?